If NaCl is doped with 10^-4 mol % of SrCl₂, the concentration of cation vacancies will be

(N_A = 6.023 x 10²³ mol^-1)                                                                             

1. 6.023 x 10¹⁵ mol^-1             

2. 6.023 x 10¹⁶ mol^-1

3. 6.023 x 10¹⁷ mol^-1             

4. 6.023 x 10¹⁴ mol^-1                                     

Lithium borohydride (LiBH₄), crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are:- a = 6.81 Å , b = 4.43 Å, c = 7.17 Å. If the molar mass of LiBH₄ is 21.76 g mol^-1, then the density of the crystal is:-

1. 0.67 g cm^-3

2. 0.58 g cm^-3

3. 1.23 g cm^-3

4. None of the above.

A solid has a bcc structure. If the distance of closest approach between the two atoms is 1.73 Å, then the edge length of the cell is:

1. 200 pm

2. $\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$ pm

3. 142.2 pm

4. $\sqrt{\mathrm{2 }}$pm

The anions (A) form hexagonal closest packing and atoms (C) occupy only 2/3 of octahedral voids in it. The general formula of the compound is-
1. CA
2. CA₂
3. C₂A₃
4. C₃A₂

How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1.00g?

[ Atomic masses Na=23, Cl = 35.5]

1. 2.57 x 10²¹

2. 5.14 x 10²¹

3. 1.28 x 10²¹

4. 1.71 x 10²¹

The most efficient packing of similar spheres is obtained in 

1. The simple cubic system and the body centered cubic system

2. The simple cubic system and the hexagonal close packed system

3. The face centered cubic system and the hexagonal close packed system

4. The body centered cubic system and the face centered cubic system

In a compound, atoms of element Y form ccp lattice, and those of element X occupy 2/3^rd of tetrahedral voids. The formula of the compound will be :-
1. X₃Y₄
2. X₄Y₃ 
3. X₂Y₃
4. X₂Y

Close packing is maximum in the crystal lattice of:

1. simple cubic

2. face centred

3. body centred

4. none of these

The density of KCl is 1.9893 g cm^-3 and the length of a side unit cell is 6.29082 Å as determined by X-ray diffraction. The value of Avogadro’s number calculated from this data is:-

1. 6.017 x 10²³

2. 6.023 x 10²²

3. 7.03 x 10²³

4. 6.01 x 10¹⁹

Oxygen atoms form fcc unit cells with 'A' atoms occupying all tetrahedral voids and 'B' atoms occupying all octahedral voids. If atoms are removed from two of the body diagonals then the formula of the resultant compound formed is: 

1. A₄B₄O₇

2. A₈B₆O₇

3. A₈B₈O₇

4. A₆B₈O₆